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  1. #1
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    supremum

    Hello. I need to find the supremum and infimum of the set S={sqrt(n) - [n] : n in N the natural numbers}. [n] stands for the greatest integer m in Z such that m is less than or equal to n.

    I assume the infimum is zero since you can't take the square root of a negative number (and fractions between 0 and 1 will be 0 in [n]) and I'm pretty sure there is no supremum since the function is unbounded. How would I prove these (both supremum and infimum) for an epsilon greater than zero?
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  2. #2
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    I do not think that you have written what you meant to write.
    If n is a natural number [n]=n. Is that what you meant.
    Or did you mean something like \left[ {\sqrt n  - n} \right]?
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  3. #3
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    Quote Originally Posted by jmarshall View Post
    Hello. I need to find the supremum and infimum of the set S={sqrt(n) - [n] : n in N the natural numbers}. [n] stands for the greatest integer m in Z such that m is less than or equal to n.

    I assume the infimum is zero since you can't take the square root of a negative number (and fractions between 0 and 1 will be 0 in [n]) and I'm pretty sure there is no supremum since the function is unbounded. How would I prove these (both supremum and infimum) for an epsilon greater than zero?
    supS = 0 and infS = - \infty

    i'm not sure what you are talking about with the epsilon greater than zero thing. that is not something we use for supremums and infimums, at least not that i've seen. are you sure you're not talking about some limit here?

    i agree with what Plato said. something is off here. (which means my answer may be wrong, i think. make sure you have the right question so i can modify it)
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    the epsilon greater than zero thing. that is not something we use for supremums and infimums, at least not that i've seen. are you sure you're not talking about some limit here?
    Actually we use an epsilon-method to prove that some number is the supremum of a set.
    To prove that a = \sup (S) we show that a is an upper bound of S and \left( {\forall \varepsilon  > 0} \right)\left( {\exists b \in S} \right)\left[ {a - \varepsilon  < b \le a} \right].
    If we can show that then a = \sup (S).
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    Actually we use an epsilon-method to prove that some number is the supremum of a set.
    To prove that a = \sup (S) we show that a is an upper bound of S and \left( {\forall \varepsilon > 0} \right)\left( {\exists b \in S} \right)\left[ {a - \varepsilon < b \le a} \right].
    If we can show that then a = \sup (S).
    Ah! that does look familiar! Thank you. I guess some review is in order for me
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  6. #6
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    You're right. I didn't write it correctly.

    S = {sqrt(n) - [sqrt(n)]: n in N}

    Can someone show me how to find the supremum and infimum using the epsilon notation?
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  7. #7
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    I will write greatest integer function as \left\lfloor x \right\rfloor .
    Now \left( {\forall x} \right)\left[ {\left\lfloor x \right\rfloor  \le x < \left\lfloor x \right\rfloor  + 1} \right] which implies that \left( {\forall n \in N} \right)\left[ {0 \le \sqrt n  - \left\lfloor {\sqrt n } \right\rfloor  <  + 1} \right].
    It is clear that n = 1\quad  \Rightarrow \quad \sqrt n  - \left\lfloor {\sqrt n } \right\rfloor  = 0. Moreover, that means that 0 is a lower bound that is in the set so 0 = \inf \left\{ {\sqrt n  - \left\lfloor {\sqrt n } \right\rfloor :n \in N} \right\}.

    It is also clear that 1 is an upper bound for the set \left\{ {\sqrt n  - \left\lfloor {\sqrt n } \right\rfloor :n \in N} \right\}.
    Can you prove 1 = \sup \left\{ {\sqrt n  - \left\lfloor {\sqrt n } \right\rfloor :n \in N} \right\}?
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  8. #8
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    Well, how do you prove the upper bound for all n? With the infimum, it seems you just randomly picked n = 1. I understand the supremum is one, but how do you show it rather than just picking numbers and saying "yeah it's one"?
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  9. #9
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    It is not at all clear that you are ready for this problem!
    First of all, no one simply picked values of n.
    I showed that a lower bound belonged to the set.
    In that case that value is inf of the set.

    However, you will not ever find a value of n for which there is an upper bound that belongs to the set.
    You must show that 1 is an upper bound and is also the least upper bound.
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  10. #10
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    OK, I guess I see what you're doing. You showed that 0 is a lower bound and 1 is an upper bound. I guess I don't know how to show that 1 is a part of the set since you can't pick values that make the function equal one. You can for zero, but you can't for one. That's correct, right? So obviously this is a little bit tougher to show and I don't know how to.

    It seems completely different from the first one (the lower bound) since one isn't part of the set; it's just an upper bound. How do I prove that it's the lowest upper bound?
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